Integrand size = 111, antiderivative size = 25 \[ \int \frac {1}{8} \left (5 x+15 x^2+10 x^3+e^{2 x} \left (30 x+45 x^2\right )+e^x \left (-25 x-55 x^2-15 x^3\right )+\left (e^{2 x} \left (-25 x-30 x^2\right )+e^x \left (10 x+20 x^2+5 x^3\right )\right ) \log (2 x)+e^{2 x} \left (5 x+5 x^2\right ) \log ^2(2 x)\right ) \, dx=\frac {5}{16} x^2 \left (1+x-e^x (3-\log (2 x))\right )^2 \]
Time = 0.08 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {1}{8} \left (5 x+15 x^2+10 x^3+e^{2 x} \left (30 x+45 x^2\right )+e^x \left (-25 x-55 x^2-15 x^3\right )+\left (e^{2 x} \left (-25 x-30 x^2\right )+e^x \left (10 x+20 x^2+5 x^3\right )\right ) \log (2 x)+e^{2 x} \left (5 x+5 x^2\right ) \log ^2(2 x)\right ) \, dx=\frac {5}{16} x^2 \left (1-3 e^x+x+e^x \log (2 x)\right )^2 \]
Integrate[(5*x + 15*x^2 + 10*x^3 + E^(2*x)*(30*x + 45*x^2) + E^x*(-25*x - 55*x^2 - 15*x^3) + (E^(2*x)*(-25*x - 30*x^2) + E^x*(10*x + 20*x^2 + 5*x^3) )*Log[2*x] + E^(2*x)*(5*x + 5*x^2)*Log[2*x]^2)/8,x]
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {1}{8} \left (10 x^3+15 x^2+e^{2 x} \left (45 x^2+30 x\right )+e^{2 x} \left (5 x^2+5 x\right ) \log ^2(2 x)+e^x \left (-15 x^3-55 x^2-25 x\right )+\left (e^{2 x} \left (-30 x^2-25 x\right )+e^x \left (5 x^3+20 x^2+10 x\right )\right ) \log (2 x)+5 x\right ) \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{8} \int \left (10 x^3+15 x^2+5 x+5 e^{2 x} \left (x^2+x\right ) \log ^2(2 x)+15 e^{2 x} \left (3 x^2+2 x\right )-5 e^x \left (3 x^3+11 x^2+5 x\right )-5 \left (e^{2 x} \left (6 x^2+5 x\right )-e^x \left (x^3+4 x^2+2 x\right )\right ) \log (2 x)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{8} \left (5 \int e^{2 x} x^2 \log ^2(2 x)dx+5 \int e^{2 x} x \log ^2(2 x)dx+\frac {5 \operatorname {ExpIntegralEi}(2 x)}{4}+\frac {5 x^4}{2}-15 e^x x^3+5 x^3+5 e^x x^3 \log (2 x)-15 e^x x^2+\frac {45}{2} e^{2 x} x^2+\frac {5 x^2}{2}+5 e^x x^2 \log (2 x)-15 e^{2 x} x^2 \log (2 x)-\frac {5 e^{2 x}}{4}+\frac {5}{2} e^{2 x} x \log (2 x)-\frac {5}{4} e^{2 x} \log (2 x)\right )\) |
Int[(5*x + 15*x^2 + 10*x^3 + E^(2*x)*(30*x + 45*x^2) + E^x*(-25*x - 55*x^2 - 15*x^3) + (E^(2*x)*(-25*x - 30*x^2) + E^x*(10*x + 20*x^2 + 5*x^3))*Log[ 2*x] + E^(2*x)*(5*x + 5*x^2)*Log[2*x]^2)/8,x]
3.3.9.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Leaf count of result is larger than twice the leaf count of optimal. \(89\) vs. \(2(22)=44\).
Time = 0.25 (sec) , antiderivative size = 90, normalized size of antiderivative = 3.60
method | result | size |
default | \(\frac {45 \,{\mathrm e}^{2 x} x^{2}}{16}-\frac {15 \,{\mathrm e}^{2 x} \ln \left (2 x \right ) x^{2}}{8}+\frac {5 \,{\mathrm e}^{2 x} \ln \left (2 x \right )^{2} x^{2}}{16}-\frac {15 \,{\mathrm e}^{x} x^{2}}{8}-\frac {15 \,{\mathrm e}^{x} x^{3}}{8}+\frac {5 \ln \left (2 x \right ) {\mathrm e}^{x} x^{2}}{8}+\frac {5 \ln \left (2 x \right ) {\mathrm e}^{x} x^{3}}{8}+\frac {5 x^{2}}{16}+\frac {5 x^{3}}{8}+\frac {5 x^{4}}{16}\) | \(90\) |
risch | \(\frac {45 \,{\mathrm e}^{2 x} x^{2}}{16}-\frac {15 \,{\mathrm e}^{2 x} \ln \left (2 x \right ) x^{2}}{8}+\frac {5 \,{\mathrm e}^{2 x} \ln \left (2 x \right )^{2} x^{2}}{16}-\frac {15 \,{\mathrm e}^{x} x^{2}}{8}-\frac {15 \,{\mathrm e}^{x} x^{3}}{8}+\frac {5 \ln \left (2 x \right ) {\mathrm e}^{x} x^{2}}{8}+\frac {5 \ln \left (2 x \right ) {\mathrm e}^{x} x^{3}}{8}+\frac {5 x^{2}}{16}+\frac {5 x^{3}}{8}+\frac {5 x^{4}}{16}\) | \(90\) |
parallelrisch | \(\frac {45 \,{\mathrm e}^{2 x} x^{2}}{16}-\frac {15 \,{\mathrm e}^{2 x} \ln \left (2 x \right ) x^{2}}{8}+\frac {5 \,{\mathrm e}^{2 x} \ln \left (2 x \right )^{2} x^{2}}{16}-\frac {15 \,{\mathrm e}^{x} x^{2}}{8}-\frac {15 \,{\mathrm e}^{x} x^{3}}{8}+\frac {5 \ln \left (2 x \right ) {\mathrm e}^{x} x^{2}}{8}+\frac {5 \ln \left (2 x \right ) {\mathrm e}^{x} x^{3}}{8}+\frac {5 x^{2}}{16}+\frac {5 x^{3}}{8}+\frac {5 x^{4}}{16}\) | \(90\) |
int(1/8*(5*x^2+5*x)*exp(x)^2*ln(2*x)^2+1/8*((-30*x^2-25*x)*exp(x)^2+(5*x^3 +20*x^2+10*x)*exp(x))*ln(2*x)+1/8*(45*x^2+30*x)*exp(x)^2+1/8*(-15*x^3-55*x ^2-25*x)*exp(x)+5/4*x^3+15/8*x^2+5/8*x,x,method=_RETURNVERBOSE)
45/16*exp(2*x)*x^2-15/8*exp(2*x)*ln(2*x)*x^2+5/16*exp(2*x)*ln(2*x)^2*x^2-1 5/8*exp(x)*x^2-15/8*exp(x)*x^3+5/8*ln(2*x)*exp(x)*x^2+5/8*ln(2*x)*exp(x)*x ^3+5/16*x^2+5/8*x^3+5/16*x^4
Leaf count of result is larger than twice the leaf count of optimal. 78 vs. \(2 (19) = 38\).
Time = 0.26 (sec) , antiderivative size = 78, normalized size of antiderivative = 3.12 \[ \int \frac {1}{8} \left (5 x+15 x^2+10 x^3+e^{2 x} \left (30 x+45 x^2\right )+e^x \left (-25 x-55 x^2-15 x^3\right )+\left (e^{2 x} \left (-25 x-30 x^2\right )+e^x \left (10 x+20 x^2+5 x^3\right )\right ) \log (2 x)+e^{2 x} \left (5 x+5 x^2\right ) \log ^2(2 x)\right ) \, dx=\frac {5}{16} \, x^{2} e^{\left (2 \, x\right )} \log \left (2 \, x\right )^{2} + \frac {5}{16} \, x^{4} + \frac {5}{8} \, x^{3} + \frac {45}{16} \, x^{2} e^{\left (2 \, x\right )} + \frac {5}{16} \, x^{2} - \frac {15}{8} \, {\left (x^{3} + x^{2}\right )} e^{x} - \frac {5}{8} \, {\left (3 \, x^{2} e^{\left (2 \, x\right )} - {\left (x^{3} + x^{2}\right )} e^{x}\right )} \log \left (2 \, x\right ) \]
integrate(1/8*(5*x^2+5*x)*exp(x)^2*log(2*x)^2+1/8*((-30*x^2-25*x)*exp(x)^2 +(5*x^3+20*x^2+10*x)*exp(x))*log(2*x)+1/8*(45*x^2+30*x)*exp(x)^2+1/8*(-15* x^3-55*x^2-25*x)*exp(x)+5/4*x^3+15/8*x^2+5/8*x,x, algorithm=\
5/16*x^2*e^(2*x)*log(2*x)^2 + 5/16*x^4 + 5/8*x^3 + 45/16*x^2*e^(2*x) + 5/1 6*x^2 - 15/8*(x^3 + x^2)*e^x - 5/8*(3*x^2*e^(2*x) - (x^3 + x^2)*e^x)*log(2 *x)
Leaf count of result is larger than twice the leaf count of optimal. 88 vs. \(2 (20) = 40\).
Time = 0.30 (sec) , antiderivative size = 88, normalized size of antiderivative = 3.52 \[ \int \frac {1}{8} \left (5 x+15 x^2+10 x^3+e^{2 x} \left (30 x+45 x^2\right )+e^x \left (-25 x-55 x^2-15 x^3\right )+\left (e^{2 x} \left (-25 x-30 x^2\right )+e^x \left (10 x+20 x^2+5 x^3\right )\right ) \log (2 x)+e^{2 x} \left (5 x+5 x^2\right ) \log ^2(2 x)\right ) \, dx=\frac {5 x^{4}}{16} + \frac {5 x^{3}}{8} + \frac {5 x^{2}}{16} + \frac {\left (40 x^{2} \log {\left (2 x \right )}^{2} - 240 x^{2} \log {\left (2 x \right )} + 360 x^{2}\right ) e^{2 x}}{128} + \frac {\left (80 x^{3} \log {\left (2 x \right )} - 240 x^{3} + 80 x^{2} \log {\left (2 x \right )} - 240 x^{2}\right ) e^{x}}{128} \]
integrate(1/8*(5*x**2+5*x)*exp(x)**2*ln(2*x)**2+1/8*((-30*x**2-25*x)*exp(x )**2+(5*x**3+20*x**2+10*x)*exp(x))*ln(2*x)+1/8*(45*x**2+30*x)*exp(x)**2+1/ 8*(-15*x**3-55*x**2-25*x)*exp(x)+5/4*x**3+15/8*x**2+5/8*x,x)
5*x**4/16 + 5*x**3/8 + 5*x**2/16 + (40*x**2*log(2*x)**2 - 240*x**2*log(2*x ) + 360*x**2)*exp(2*x)/128 + (80*x**3*log(2*x) - 240*x**3 + 80*x**2*log(2* x) - 240*x**2)*exp(x)/128
Leaf count of result is larger than twice the leaf count of optimal. 125 vs. \(2 (19) = 38\).
Time = 0.31 (sec) , antiderivative size = 125, normalized size of antiderivative = 5.00 \[ \int \frac {1}{8} \left (5 x+15 x^2+10 x^3+e^{2 x} \left (30 x+45 x^2\right )+e^x \left (-25 x-55 x^2-15 x^3\right )+\left (e^{2 x} \left (-25 x-30 x^2\right )+e^x \left (10 x+20 x^2+5 x^3\right )\right ) \log (2 x)+e^{2 x} \left (5 x+5 x^2\right ) \log ^2(2 x)\right ) \, dx=\frac {5}{16} \, x^{4} + \frac {5}{8} \, x^{3} + \frac {5}{16} \, x^{2} + \frac {5}{32} \, {\left (4 \, x^{2} {\left (\log \left (2\right ) - 3\right )} \log \left (x\right ) + 2 \, x^{2} \log \left (x\right )^{2} + 2 \, {\left (\log \left (2\right )^{2} - 6 \, \log \left (2\right )\right )} x^{2} + 6 \, x - 3\right )} e^{\left (2 \, x\right )} + \frac {15}{32} \, {\left (6 \, x^{2} - 2 \, x + 1\right )} e^{\left (2 \, x\right )} + \frac {5}{8} \, {\left (x^{3} \log \left (2\right ) + x^{2} {\left (\log \left (2\right ) - 1\right )} + {\left (x^{3} + x^{2}\right )} \log \left (x\right ) + x - 1\right )} e^{x} - \frac {5}{8} \, {\left (3 \, x^{3} + 2 \, x^{2} + x - 1\right )} e^{x} \]
integrate(1/8*(5*x^2+5*x)*exp(x)^2*log(2*x)^2+1/8*((-30*x^2-25*x)*exp(x)^2 +(5*x^3+20*x^2+10*x)*exp(x))*log(2*x)+1/8*(45*x^2+30*x)*exp(x)^2+1/8*(-15* x^3-55*x^2-25*x)*exp(x)+5/4*x^3+15/8*x^2+5/8*x,x, algorithm=\
5/16*x^4 + 5/8*x^3 + 5/16*x^2 + 5/32*(4*x^2*(log(2) - 3)*log(x) + 2*x^2*lo g(x)^2 + 2*(log(2)^2 - 6*log(2))*x^2 + 6*x - 3)*e^(2*x) + 15/32*(6*x^2 - 2 *x + 1)*e^(2*x) + 5/8*(x^3*log(2) + x^2*(log(2) - 1) + (x^3 + x^2)*log(x) + x - 1)*e^x - 5/8*(3*x^3 + 2*x^2 + x - 1)*e^x
Leaf count of result is larger than twice the leaf count of optimal. 141 vs. \(2 (19) = 38\).
Time = 0.29 (sec) , antiderivative size = 141, normalized size of antiderivative = 5.64 \[ \int \frac {1}{8} \left (5 x+15 x^2+10 x^3+e^{2 x} \left (30 x+45 x^2\right )+e^x \left (-25 x-55 x^2-15 x^3\right )+\left (e^{2 x} \left (-25 x-30 x^2\right )+e^x \left (10 x+20 x^2+5 x^3\right )\right ) \log (2 x)+e^{2 x} \left (5 x+5 x^2\right ) \log ^2(2 x)\right ) \, dx=\frac {5}{16} \, x^{2} e^{\left (2 \, x\right )} \log \left (2 \, x\right )^{2} + \frac {5}{16} \, x^{4} + \frac {5}{8} \, x^{3} - \frac {5}{8} \, x^{2} e^{x} - \frac {5}{32} \, {\left (2 \, x - 1\right )} e^{\left (2 \, x\right )} \log \left (2 \, x\right ) + \frac {5}{16} \, x^{2} + \frac {15}{32} \, {\left (6 \, x^{2} - 2 \, x + 1\right )} e^{\left (2 \, x\right )} + \frac {15}{16} \, x e^{\left (2 \, x\right )} - \frac {5}{8} \, {\left (3 \, x^{3} + 2 \, x^{2} + x - 1\right )} e^{x} + \frac {5}{8} \, x e^{x} - \frac {5}{32} \, {\left ({\left (12 \, x^{2} - 2 \, x + 1\right )} e^{\left (2 \, x\right )} - 4 \, {\left (x^{3} + x^{2}\right )} e^{x}\right )} \log \left (2 \, x\right ) - \frac {15}{32} \, e^{\left (2 \, x\right )} - \frac {5}{8} \, e^{x} \]
integrate(1/8*(5*x^2+5*x)*exp(x)^2*log(2*x)^2+1/8*((-30*x^2-25*x)*exp(x)^2 +(5*x^3+20*x^2+10*x)*exp(x))*log(2*x)+1/8*(45*x^2+30*x)*exp(x)^2+1/8*(-15* x^3-55*x^2-25*x)*exp(x)+5/4*x^3+15/8*x^2+5/8*x,x, algorithm=\
5/16*x^2*e^(2*x)*log(2*x)^2 + 5/16*x^4 + 5/8*x^3 - 5/8*x^2*e^x - 5/32*(2*x - 1)*e^(2*x)*log(2*x) + 5/16*x^2 + 15/32*(6*x^2 - 2*x + 1)*e^(2*x) + 15/1 6*x*e^(2*x) - 5/8*(3*x^3 + 2*x^2 + x - 1)*e^x + 5/8*x*e^x - 5/32*((12*x^2 - 2*x + 1)*e^(2*x) - 4*(x^3 + x^2)*e^x)*log(2*x) - 15/32*e^(2*x) - 5/8*e^x
Time = 12.76 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {1}{8} \left (5 x+15 x^2+10 x^3+e^{2 x} \left (30 x+45 x^2\right )+e^x \left (-25 x-55 x^2-15 x^3\right )+\left (e^{2 x} \left (-25 x-30 x^2\right )+e^x \left (10 x+20 x^2+5 x^3\right )\right ) \log (2 x)+e^{2 x} \left (5 x+5 x^2\right ) \log ^2(2 x)\right ) \, dx=\frac {5\,x^2\,{\left (x-3\,{\mathrm {e}}^x+\ln \left (2\,x\right )\,{\mathrm {e}}^x+1\right )}^2}{16} \]